1. ## 2e^2.1

2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ?

2. ## Re: 2e^2.1

Hint: \displaystyle \begin{align*} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \end{align*}

Substitute \displaystyle \begin{align*} x = 2.1 \end{align*} and evaluate to as much accuracy as you desire.

3. ## Re: 2e^2.1

Originally Posted by mathsgirl123431
2e^2.1

that's two e to the power of 2.1

does this equal 16.33 (2dp) ?
Yes.

Originally Posted by Prove It
Hint: \displaystyle \begin{align*} e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \dots \end{align*}

Substitute \displaystyle \begin{align*} x = 2.1 \end{align*} and evaluate to as much accuracy as you desire.
To calculate $e^{2.1}$ to 2 decimal places would require computing up to $\frac{x^9}{9!}$.

4. ## Re: 2e^2.1

Thanks guys, I appreciate the help

5. ## Re: 2e^2.1

If it were me, I would use a calculator!

6. ## Re: 2e^2.1

Here's another perfectly fine way of doing this. You know how to take integer powers and suppose you have $e$ handy, to good precision. What can we do then? We have

$2e^{2.1} = 2e^2 e^{1/10}$

Now you can use the Taylor series for that last exponential

$e^{1/h} = 1+\frac{1}{h}+\frac{1}{2h^2}+\frac{1}{6}h^3 + \cdots$

If you have $2e^2$ to several decimal places in precision, then you can get away with using less terms in this series. It's not a big save but it's kind of neat.

7. ## Re: 2e^2.1

Originally Posted by HallsofIvy
If it were me, I would use a calculator!
YES!
Like, why use the arithmetic series formula to add 1+2+3 ?

8. ## Re: 2e^2.1

I have finished up with the following and I want to find t;

2.6 = (e^-0.2t) + (e^-0.5t)

Any help and pointers would be appreciated. Thank you

9. ## Re: 2e^2.1

Denote x = e^(-0.1t). Then your equation is 2.6 = x^2 + x^5. This will probably have to be solved numerically.